A line l passing through the origin is perpendicular to the lines \,l_1:( 3 + t )\widehat i + ( - 1 + 2t )\widehat j + ( 4 + 2t )\widehat k,\,\,\,\,\, - \infty \lt t \lt \infty l_2:( 3 + 2s )\widehat i + ( 3 + 2s )\widehat j + ( 2 + s )\widehat k,\,\,\,\,\, - \infty \lt s \lt \infty Then, the coordinate(s) of the points(s) on l_2 at a distance of \sqrt 17 from the point of intersection of l and l_1 is (are)

( - 1, - 1,0 ), ( 7 \over 9,7 \over 9,8 \over 9 )

( 7 \over 3,7 \over 3,5 \over 3 )

( 7 \over 9,7 \over 9,8 \over 9 )

3D Geometry PYQ — JEE Advanced 2013 · Shift 1 · Q39 with Solution

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Correct answer: B, D

Write the lines in parametric/vector form

Given $l_1:(3+t)\hat i+(-1+2t)\hat j+(4+2t)\hat k$ so a general point on $l_1$ is $P_1(t)=(3+t,-1+2t,4+2t)$ and its direction vector is $\vec d_1=(1,2,2).$

Similarly, $l_2:(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k$ so a general point on $l_2$ is $P_2(s)=(3+2s,3+2s,2+s)$ and its direction vector is $\vec d_2=(2,2,1).$

Find the direction of line $l$

The line $l$ passes through the origin and is perpendicular to both $l_1$ and $l_2$ . Hence its direction vector must be perpendicular to both $\vec d_1$ and $\vec d_2$ . So, $\vec d=\vec d_1\times \vec d_2.$

Compute:

\begin{vmatrix} \hat i&\hat j&\hat k\\ 1&2&2\\ 2&2&1 \end{vmatrix}$$ $$=\hat i(2\cdot1-2\cdot2)-\hat j(1\cdot1-2\cdot2)+\hat k(1\cdot2-2\cdot2)$$ $$=(-2,3,-2).$$ Thus line $l$ is $$\vec r=\lambda(-2,3,-2).$$ 3. **Find the intersection point of $l$ and $l_1$** Let the intersection point satisfy $$(3+t,-1+2t,4+2t)=\lambda(-2,3,-2).$$ So, \begin{align*} 3+t&=-2\lambda \quad ...(1)\\ -1+2t&=3\lambda \quad ...(2)\\ 4+2t&=-2\lambda \quad ...(3) \end{align*} From (1) and (3), since both equal $-2\lambda$, $$3+t=4+2t$$ which gives $$t=-1.$$ Then from (1), $$3-1=2=-2\lambda \Rightarrow \lambda=-1.$$ Hence the intersection point is $$Q=P_1(-1)=(2,-3,2).$$ 4. **Find points on $l_2$ at distance $\sqrt{17}$ from $Q$** A general point on $l_2$ is $$P(s)=(3+2s,3+2s,2+s).$$ We need $$|P(s)-Q|=\sqrt{17}.$$ Now, $$P(s)-Q=(3+2s-2,\;3+2s-(-3),\;2+s-2)=(1+2s,6+2s,s).$$ So, $$|(1+2s,6+2s,s)|^2=17.$$ That is, $$(1+2s)^2+(6+2s)^2+s^2=17.$$ Expand: \begin{align*} (1+2s)^2&=1+4s+4s^2,\\ (6+2s)^2&=36+24s+4s^2. \end{align*} Thus, $$1+4s+4s^2+36+24s+4s^2+s^2=17$$ $$9s^2+28s+37=17$$ $$9s^2+28s+20=0.$$ Solve: $$9s^2+28s+20=(9s+10)(s+2)=0.$$ So, $$s=-\frac{10}{9}\quad \text{or}\quad s=-2.$$ 5. **Find the corresponding points on $l_2$** - For $s=-2$: $$P(-2)=(3-4,3-4,2-2)=(-1,-1,0).$$ This is **Option B**. - For $s=-\frac{10}{9}$: $$P\left(-\frac{10}{9}\right)=\left(3-\frac{20}{9},3-\frac{20}{9},2-\frac{10}{9}\right) =\left(\frac{7}{9},\frac{7}{9},\frac{8}{9}\right).$$ This is **Option D**. 6. **Check all options** - **A:** $\left(\frac73,\frac73,\frac53\right)$ is not obtained for any valid $s$. - **B:** $(-1,-1,0)$ is correct. - **C:** $(1,1,1)$ is not obtained for any valid $s$. - **D:** $\left(\frac79,\frac79,\frac89\right)$ is correct. Therefore, the correct options are $$\boxed{\text{B, D}}.$$

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View written solutionFree

Correct answer: B, D

Write the lines in parametric/vector form

Given $l_1:(3+t)\hat i+(-1+2t)\hat j+(4+2t)\hat k$ so a general point on $l_1$ is $P_1(t)=(3+t,-1+2t,4+2t)$ and its direction vector is $\vec d_1=(1,2,2).$

Similarly, $l_2:(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k$ so a general point on $l_2$ is $P_2(s)=(3+2s,3+2s,2+s)$ and its direction vector is $\vec d_2=(2,2,1).$

Find the direction of line $l$

Compute:

\begin{vmatrix} \hat i&\hat j&\hat k\\ 1&2&2\\ 2&2&1 \end{vmatrix}$$ $$=\hat i(2\cdot1-2\cdot2)-\hat j(1\cdot1-2\cdot2)+\hat k(1\cdot2-2\cdot2)$$ $$=(-2,3,-2).$$ Thus line $l$ is $$\vec r=\lambda(-2,3,-2).$$ 3. **Find the intersection point of $l$ and $l_1$** Let the intersection point satisfy $$(3+t,-1+2t,4+2t)=\lambda(-2,3,-2).$$ So, \begin{align*} 3+t&=-2\lambda \quad ...(1)\\ -1+2t&=3\lambda \quad ...(2)\\ 4+2t&=-2\lambda \quad ...(3) \end{align*} From (1) and (3), since both equal $-2\lambda$, $$3+t=4+2t$$ which gives $$t=-1.$$ Then from (1), $$3-1=2=-2\lambda \Rightarrow \lambda=-1.$$ Hence the intersection point is $$Q=P_1(-1)=(2,-3,2).$$ 4. **Find points on $l_2$ at distance $\sqrt{17}$ from $Q$** A general point on $l_2$ is $$P(s)=(3+2s,3+2s,2+s).$$ We need $$|P(s)-Q|=\sqrt{17}.$$ Now, $$P(s)-Q=(3+2s-2,\;3+2s-(-3),\;2+s-2)=(1+2s,6+2s,s).$$ So, $$|(1+2s,6+2s,s)|^2=17.$$ That is, $$(1+2s)^2+(6+2s)^2+s^2=17.$$ Expand: \begin{align*} (1+2s)^2&=1+4s+4s^2,\\ (6+2s)^2&=36+24s+4s^2. \end{align*} Thus, $$1+4s+4s^2+36+24s+4s^2+s^2=17$$ $$9s^2+28s+37=17$$ $$9s^2+28s+20=0.$$ Solve: $$9s^2+28s+20=(9s+10)(s+2)=0.$$ So, $$s=-\frac{10}{9}\quad \text{or}\quad s=-2.$$ 5. **Find the corresponding points on $l_2$** - For $s=-2$: $$P(-2)=(3-4,3-4,2-2)=(-1,-1,0).$$ This is **Option B**. - For $s=-\frac{10}{9}$: $$P\left(-\frac{10}{9}\right)=\left(3-\frac{20}{9},3-\frac{20}{9},2-\frac{10}{9}\right) =\left(\frac{7}{9},\frac{7}{9},\frac{8}{9}\right).$$ This is **Option D**. 6. **Check all options** - **A:** $\left(\frac73,\frac73,\frac53\right)$ is not obtained for any valid $s$. - **B:** $(-1,-1,0)$ is correct. - **C:** $(1,1,1)$ is not obtained for any valid $s$. - **D:** $\left(\frac79,\frac79,\frac89\right)$ is correct. Therefore, the correct options are $$\boxed{\text{B, D}}.$$

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